Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.7$ years; the standard deviation is $3.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $3.5$ years.
Answer: $13.7$ $10.3$ $17.1$ $6.9$ $20.5$ $3.5$ $23.9$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $13.7$ years. We know the standard deviation is $3.4$ years, so one standard deviation below the mean is $10.3$ years and one standard deviation above the mean is $17.1$ years. Two standard deviations below the mean is $6.9$ years and two standard deviations above the mean is $20.5$ years. Three standard deviations below the mean is $3.5$ years and three standard deviations above the mean is $23.9$ years. We are interested in the probability of a lion living less than $3.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $3.5$ years and the other half $({0.15\%})$ will live longer than $23.9$ years. The probability of a particular lion living less than $3.5$ years is ${0.15\%}$.